ne formulation mkm

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DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING, MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL. Saturday, October 3, 2015 1 ICE 425 Presentation Created By: MUKUND KUMAR MENON, ASSISTANT PROFESSOR, DEPT. OF INSTRUMENTATION & CONTROL ENGG., MIT, MANIPAL. R obotic S ystems & C ontrol NEWTON EULER {NE } FORMULATION NEWTON EULER {NE} FORMULATION

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NE Formulation MKM

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Page 1: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Saturday, October 3, 2015 1

ICE 425

Presentation Created By:MUKUND KUMAR MENON,ASSISTANT PROFESSOR,DEPT. OF INSTRUMENTATION & CONTROL ENGG.,MIT, MANIPAL.

Robotic Systems & Control

NEWTON EULER {NE} FORMULATION

NEWTON EULER {NE} FORMULATION

Page 2: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

αc,i = acceleration of the centre of mass of link i

αe,i = acceleration of the end of link i (which is the origin of frame i + 1)

ωi = angular velocity of frame i with respect to frame 0

αi = angular acceleration of frame i with respect to frame 0

zi = axis of actuation of frame i with respect to frame 0

gi = acceleration due to gravity

fi = force exerted by link (i − 1) on link i

τi = torque exerted by link (i − 1) on link i

LET

Saturday, October 3, 2015 2NEWTON EULER {NE} FORMULATION

Page 3: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Rii+1 = rotation matrix from frame ‘i’ to frame ‘i + 1’

mi = the mass of link i

Ii = inertia-tensor of link i about a frame parallel to frame ‘i’, whose origin is

at the centre of mass of link ‘i'

r i−1,ci = vector from the origin of frame (i − 1) to the centre of mass of link i

r i−1,i = vector from the origin of frame (i − 1) to the origin of frame i

r i,ci = vector from the origin of frame i to the centre of mass of link i

LET

Saturday, October 3, 2015 3NEWTON EULER {NE} FORMULATION

Page 4: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Fig: Forces and torques acting on a random link ‘i’

Saturday, October 3, 2015 4NEWTON EULER {NE} FORMULATION

By the law of action and reaction, f i is the force exerted by link i − 1 on link i, and −f i+1 is the forceexerted by link i+1 on link i. According to the definitions above, fi is expressed in frame i while −f i+1 isexpressed in frame i + 1. Thus, to express both forces in frame i, it is required to post-multiply the latterwith Ri

i+1 . The same apply to the torque, again by the law of action and reaction.

Page 5: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Saturday, October 3, 2015 5

1. FORCE BALANCE EQUATION:

1 1 ,

i

i i i i i i c if m g R f m a

, 1 1

i

i i c i i i i if m a R f m g

At link ‘i’:

NEWTON EULER {NE} FORMULATION

c

link

f ma

Page 6: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Saturday, October 3, 2015 6

First, the moment exerted by a force ‘f’ about a point is given by (f × r), where

‘r’ is the radial vector from the point where the force is applied to the point

where the moment is computed.

Second, the vector migi does not appear in the moment balance since it is

applied directly at the centre of mass.

2. MOMENT BALANCE EQUATION:

NEWTON EULER {NE} FORMULATION

Page 7: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Saturday, October 3, 2015 7NEWTON EULER {NE} FORMULATION

link

I I

1, 1 1 1 1 ,

i i

i i i ci i i i i i ci i i i i if r R R f r I I

1, 1 1 1 1 ,

i i

i i i i i i i i ci i i i i i ciI I f r R R f r

Page 8: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Saturday, October 3, 2015 NEWTON EULER {NE} FORMULATION 8

Because of the fact that the angular velocity of frame i equals the angular

velocity of frame ‘i−1’ PLUS the added rotation from joint i, using rotation

matrices this leads to:

the rotation of joint i expressedin frame i

NOW,

Page 9: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Saturday, October 3, 2015 NEWTON EULER {NE} FORMULATION 9

It is interesting & also vitally important to note that:

which means

ii

αi is the derivative of

the angular velocity of

frame i, but expressed

in frame i.

Page 10: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Saturday, October 3, 2015 NEWTON EULER {NE} FORMULATION 10

The time derivative of:

becomes:

and expressed in frame ‘i', it directly becomes:

Page 11: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Saturday, October 3, 2015 NEWTON EULER {NE} FORMULATION 11

Now it only remains to find an expression for ac,i .

The linear velocity of the centre of mass of link i is expressed as:

constant in frame i

Differentiating both sides, we get:

Page 12: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Saturday, October 3, 2015 NEWTON EULER {NE} FORMULATION 12

Multiplying with rotation matrices and using the fact that:

Then, the final expression for the acceleration of the centre of mass of link i,

expressed in frame i, becomes:

and,

THE ACCELERATION OF THE END OF THE LINK,

Page 13: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Saturday, October 3, 2015 NEWTON EULER {NE} FORMULATION 13

This completes the recursive formulation, and The Newton-Euler

formulation of an n-link manipulator can be stated as follows :

2. Forward recursion:

1. Start with the initial conditions: ω0 = α0 = ac,0 = ae,0 = 0, fn+1 = τn+1 = 0

For increasing i, from 1 to n, solve in the following order:

Page 14: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Saturday, October 3, 2015 NEWTON EULER {NE} FORMULATION 14

3. Backward recursion:

For decreasing i, from n to 1, solve in the following order:

1,

1 1 1 1 ,

i i i i i i i i ci

i i

i i i i i ci

I I f r

R R f r

, 1 1

i

i i c i i i i if m a R f m g

Page 15: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Saturday, October 3, 2015 NEWTON EULER {NE} FORMULATION 15

Qn:Analyze the dynamics of the planar elbow manipulator using NE RecursiveMethod.

Hint:Pg 222-224, Sect.6.7, Mark W. Spong, Seth Hutchinson,and M. Vidyasagar , “Robot Modeling and Control”, FirstEdition, Wiley.

Page 16: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Saturday, October 3, 2015 NEWTON EULER {NE} FORMULATION 16

REFERENCE:

1. Herman Høifødt, “Dynamic Modeling and Simulation of Robot

Manipulators: The Newton-Euler Formulation”, Thesis, Master of

Science in Engineering Cybernetics, Norwegian University of Science

and Technology, Department of Engineering Cybernetics, June 2011.

2. Mark W. Spong, Seth Hutchinson, and M. Vidyasagar , “Robot

Modeling and Control”, First Edition, Wiley.

3. John r. Taylor, “Classical mechanics”, University Science Books, 2005.

Page 17: NE Formulation MKM

DEPARTMENT OF INSTRUMENTATION AND CONTROL ENGINEERING,

MANIPAL INSTITUTE OF TECHNOLOGY, MANIPAL.

Saturday, October 3, 2015 NEWTON EULER {NE} FORMULATION 17